Answer
$-\frac{1}{3}$
Work Step by Step
If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x$ and every real number $y$.
Therefore, $log_{8}\frac{1}{2}=-\frac{1}{3}$.
We know this because $8^{-\frac{1}{3}}=\frac{1}{8^{\frac{1}{3}}}=\frac{1}{\sqrt[3] 8}=\frac{1}{2}$ (where $\sqrt[3] 8=2$, because $2^{3}=8$).
